Answer
Vertices: $(±1,0)$
Eccentricity: $e=\frac{2\sqrt 2}{3}$
Work Step by Step
$x^2+9y^2=1$
$\frac{x^2}{1}+\frac{y^2}{\frac{1}{9}}=1$
$\frac{x^2}{1^2}+\frac{y^2}{(\frac{1}{3})^2}=1$
The major axis is horizontal.
Standard form when major axis is horizontal:
$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$
So: $a=1$ and $b=\frac{1}{3}$
$a^2=b^2+c^2$
$c^2=a^2-b^2=1-\frac{1}{9}=\frac{8}{9}$
$c=\frac{2\sqrt 2}{3}$
$e=\frac{c}{a}=\frac{\frac{2\sqrt 2}{3}}{1}=\frac{2\sqrt 2}{3}$
Vertices when major axis is horizontal:
$(±a,0)=(±1,0)$
